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Math Help - Solving This Integral

  1. #1
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    Solving This Integral

    I'm really having trouble understanding how to solve the following integrals, hopefully someone can help me

    ∫ (3/(x^2 - 7x +12))

    I'm just not sure even how to start. I know that I can take out the constant and make it:

    3 ∫ (1/(x^2 - 7x +12))

    But what should I do after that, I know that my answer will involve utilizing ln
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  2. #2
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    Hi

    You can start by rewriting
    \frac{1}{x^2-7x+12} in "simple elements"

    using the fact that x^2-7x+12 = (x-4)(x-3)

    \frac{1}{x^2-7x+12} = \frac{1}{x-4} - \frac{1}{x-3}

    \int \frac{3}{x^2-7x+12} \, dx = 3 \int (\frac{1}{x-4} - \frac{1}{x-3}) \, dx = 3\; ln\frac{|x-4|}{|x-3|}
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    You can start by rewriting
    \frac{1}{x^2-7x+12} in "simple elements"

    using the fact that x^2-7x+12 = (x-4)(x-3)

    \frac{1}{x^2-7x+12} = \frac{1}{x-4} - \frac{1}{x-3}

    \int \frac{3}{x^2-7x+12} \, dx = 3 \int (\frac{1}{x-4} - \frac{1}{x-3}) \, dx = 3\; ln\frac{|x-4|}{|x-3|}
    Thanks! I had you until this part:
    \frac{1}{x^2-7x+12} = \frac{1}{x-4} - \frac{1}{x-3}

    How did you get that?
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  4. #4
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    Quote Originally Posted by trigoon View Post
    Thanks! I had you until this part:
    \frac{1}{x^2-7x+12} = \frac{1}{x-4} - \frac{1}{x-3}

    How did you get that?
    Experience ...

    This is the nominal way to "decompose in simple elements" (translation from French, I do not know if it is called this way in English)

    \frac{1}{(x-a)(x-b)} = \frac{1}{a-b} \,(\frac{1}{x-a} - \frac{1}{x-b})
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  5. #5
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    Thanks a lot this really helps!
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